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iren2701 [21]
4 years ago
9

An green hoop with mass mh = 2.8 kg and radius rh = 0.13 m hangs from a string that goes over a blue solid disk pulley with mass

md = 2.1 kg and radius rd = 0.1 m. the other end of the string is attached to a massless axel through the center of an orange sphere on a flat horizontal surface that rolls without slipping and has mass ms = 4.3 kg and radius rs = 0.18 m. the system is released from rest. 1) what is magnitude of the linear acceleration of the hoop?
Physics
1 answer:
Otrada [13]4 years ago
7 0
The only force on the system is the mass of the hoop F net = 2.8kg*9.81m/s^2 = 27.468 N The mass equal of the rolling sphere is found by: the sphere rotates around the contact point with the table. 
So by applying the theorem of parallel axes, the moment of inertia of the sphere is computed by:I = 2/5*mR^2 for rotation about the center of mass + mR^2 for the distance of the axis of rotation from the center of mass of the sphere. 
I = 7/5*mR^2 M = 7/5*m 
Therefore, linear acceleration is computed by:F/m = 27.468 / (2.8 + 1/2*2 + 7/5*4) = 27.468/9.4 = 2.922 m/s^2 
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The force component along the displacement varies with the magnitude of the displacement, as shown in the graph. (a) 0 to 1.0 m,
Lina20 [59]
Work is force*displacement if the force and displacement is parallel. 


a. You can average the force over the distance so W = Fave*d 

<span>b The force part of that multiplication is zero. </span>

<span>c. You can form the average force for the interval from 2 to 3 and find the work for that section and then consider the interval from 3 to 4, find the work and add the 2 work results.


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5 0
3 years ago
A simple harmonic wave of wavelength 18.7 cm and amplitude 2.34 cm is propagating along a string in the negative x-direction at
storchak [24]

Answer

given,

wavelength = λ = 18.7 cm

                    = 0.187 m

amplitude , A = 2.34 cm

v = 0.38 m/s

A)  angular frequency = ?

     f = \dfrac{v}{\lambda}

     f = \dfrac{0.38}{0.187}

     f =2.03\ Hz

angular frequency ,

ω = 2π f

ω = 2π x 2.03

ω = 12.75 rad/s

B) the wave number ,

      K = \dfrac{2\pi}{\lambda}

     K= \dfrac{2\pi}{0.187}

    K =33.59\ m^{-1}

C)

as the wave is propagating in -x direction, the sign is positive between x and t

y ( x ,t) = A sin(k  x - ω t)

y ( x ,t) = 2.34  x  sin(33.59 x - 12.75 t)

4 0
3 years ago
Whats an astral projection
Dominik [7]

Answer:Esotericism

Explanation:

it’s something that’s in intentional out of body experience

5 0
3 years ago
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Which statement describes characteristics of a concave lens?
stiv31 [10]
I dont know what the statements are but concave lens are thinner in the middle which cause light to diverge or scatter
5 0
3 years ago
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Two rigid tanks of equal size and shape are filled with different gases. The tank on the left contains oxygen, and the tank on t
fredd [130]

Answer:

The number of oxygen molecules in the left container greater than the number of hydrogen molecules in the right container.

Explanation:

Given:

Molar mass of oxygen, M_O=32

Molar mass of hydrogen, M_H=2

We know ideal gas law as:

PV=nRT

where:

P = pressure of the gas

V = volume of the gas

n= no. of moles of the gas molecules

R = universal gs constant

T = temperature of the gas

∵n=\frac{m}{M}

where:

m = mass of gas in grams

M = molecular mass of the gas

∴Eq. (1) can be written as:

PV=\frac{m}{M}.RT

P=\frac{m}{V}.\frac{RT}{M}

        as: \frac{m}{V}=\rho\ (\rm density)

So,

P=\rho.\frac{RT}{M}

Now, according to given we have T,P,R same for both the gases.

P_O=P_H

\rho_O.\frac{RT}{M_O}=\rho_H.\frac{RT}{M_H}

\Rightarrow \frac{\rho_O}{32}=\frac{\rho_H}{2}

\rho_O=16\rho_H

∴The molecules of oxygen are more densely packed than the molecules of hydrogen in the same volume at the same temperature and pressure. So, <em>the number of oxygen molecules in the left container greater than the number of hydrogen molecules in the right container.</em>

5 0
3 years ago
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